I need help please with this chemistry work

If 8.6 g of ch4 and 5.9 g of o2 react, what is the mass, in grams, of h2o that is produced?

Alenkasestr [34]

This may seem confusing because they give you two masses, but all you have to do is pick one to do the calculations. Personally, I would pick O2, since the molar mass is easier to calculate. The answer would be 3.3 g (rounded for sig figs). To get this, first take the 5.9 grams of O2 and convert it to moles by dividing by the molar mass of oxygen gas, which is 32. Then, multiply both by the mole-mole ratio, which is 2:2, or simply 1:1. After that, multiply that by 18g, which is the molar mass of water to get grams of water.

REMEMBER, you have to write and balance the chemical equation before you can do any of that work.
That happens to be CH4 + 2O2 => CO2 + 2H2O

7 0

3 years ago

What happens when two positively charged objects move closer to each other?

Katena32 [7]

Answer: Objects with like charge repel each other.

6 0

3 years ago

Read 2 more answers

HEY Plzzzzzzzzzzz HELP ILL GIVE BRAINLIEST ASAPPPPPPPPPPP

vladimir1956 [14]

1. Equal
2. Properties
3. Heat
4. Reverse

3 0

2 years ago

Calculate the maximum numbers of moles and grams of H₂S that can form when 158 g of aluminum sulfide reacts with 131 g of water:

Phantasy [73]

What is Chemical Reaction?

A chemical reaction is the chemical transformation of one set of chemical components into another.

Main Content

Mass of aluminium sulfide is 158g

Mass of water is 131g

The chemical reaction: $Al_{2}S_{3} +H_{2}O _\to Al(OH)_{3} + H_{2}S$

First, balance the chemical equation

$Al_{2}S_{3} + 6H_{2}O \to 2Al(OH)_{3} + 3H_{2}S$

Aluminium sulfide has a molar mass of 150.16 g/mol and water has a molar mass of 18.02 g/mol. As a result, the moles of aluminum sulfide are computed as follows:

$n_{Al_{2}S_{3} } = \frac{Mass}{Molar mass}\\n_{Al_{2} S_{3} } = \frac{158g}{150.16g/mol} \\n_{Al_{2}S_{3} }=1.05 mol$

From the chemical reaction , the ratio of molar is 3mol $H_{2}S/1 mol Al_{2}S_{3}.$ So, the moles of hydrogen sulfide are:

$n_{H_{2} O} =\frac{131g}{18.02g/mol}$

= 7.26mol

From the chemical reaction, the molar ratio is 3 mol $H_{2}S/6$ mol $H_{2}O.$ So, the moles of hydrogen sulfide are:

Moles of $H_{2}S$ formed = 7.26 mol $H_{2}O \times \frac{3 mol H_{2}S }{6 mol H_{2} O} }$

Th liming reactant is $Al_{2}S_{3}$ beacuse the mass of $Al_{2}S_{3}$ forms less product than water. Therefore, the maximum number of moles of $H_{2}S$ is 3.15 mol. We know that molar mass of $H_{2}S$ is 34.10g/mol. So, the maximum mass of $H_{2}S$ formed is,

$m_{H_{2}S } = n_{H_{2}S } \times$ Molar mass of $H_{2}S$

= 3.15 mol $\times$ 34.10g/mol

= 107.4g

Now, multiplying the number of moles of $Al_{2}S_{3}$ by the molar ratio between $Al_2S_3$ and $H_2O$ which is 6mol $H_2O/1mol$ $Al_2S_3$ we get the number of moles of $H_2O$ reacted.